A line with slope equal to $1$ and a line with slope equal to $2$ intersect at the point $P(1,6),$ as shown. [asy]
unitsize(0.5 cm);
draw((0,-1)--(0,10),EndArrow);
draw((-10,0)--(5,0),EndArrow);
draw((-6,-1)--(5,10),linewidth(0.8));
draw((-2.5,-1)--(3,10),linewidth(0.8));
label("$x$",(5,0),E);
label("$y$",(0,10),N);
label("$P(1,6)$",(1,6),SE);
label("$Q$",(-5,0),NW);
label("$R$",(-2,0),SE);
[/asy]What is the area of $\triangle PQR?$
Answer: The slope of line segment $QP$ is $1.$ Since the "rise" of $QP$ is $6$ units, the "run" of $QP$ should also be $6$ units. Therefore, $Q$ is $6$ units horizontally to the left of $P,$ and so has coordinates $(-5,0).$

The slope of line segment $RP$ is $2.$ Since the rise of $RP$ is $6$ units, then the run of $RP$ is $\frac{1}{2}\cdot 6=3$ units. Therefore, $R$ is $3$ units horizontally to the left of $P,$ and so has coordinates $(-2,0).$

(We could have used the coordinates of $P$ and the slopes of the lines to find that the equations of the lines are $y=x+5$ and $y=2x+4$ and used them to find the coordinates of $Q$ and $R.$)

Therefore, $QR=-2-(-5)=3$ and $P$ is $6$ units above the $x$-axis. Thus, treating $QR$ as the base of $\triangle PQR,$ we find that its area is $$\frac{1}{2}\cdot 3\cdot 6=\boxed{9}.$$